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Jason Onwenu
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Homework Statement
During a test dive in 1939, prior to being accepted by the U.S. Navy, the submarine Squalus sank at a point where the depth of water was 73.0 m. The temperature at the surface was 27.0 ∘C and at the bottom it was 7.0 ∘C. The density of seawater is 1030 kg/m3. A diving bell was used to rescue 33 trapped crewmen from the Squalus. The diving bell was in the form of a circular cylinder 2.30 m high, open at the bottom and closed at the top. When the diving bell was lowered to the bottom of the sea, to what height did water rise within the diving bell? (Hint: You may ignore the relatively small variation in water pressure between the bottom of the bell and the surface of the water within the bell.)
Homework Equations
P = rho * g * height
PV = nRT
P1V1/T1 = P2V2/T2
V = pi * radius^2 * height
The Attempt at a Solution
P(bottom) = 101325 + (1030*9.8*73) = 838137 Pa
V2 = ? (The radius wasn't given...)
T2 = 280.15 K
P(top) = 101325
V1 = ? (The radius wasn't given...)
T1 = 303.15K
At this point, I couldn't really proceed with my calculations because a radius wasn't given. I'm wondering if I could solve for the radius with the information given or if the radius is even needed at all.